Question

6 multiple choice 5 points greg wrote the numbers 1 through 9 on pieces of paper and placed them in a hat. he will randomly select one piece of paper from the hat. he will not replace it. greg will then randomly select a second piece of paper from the hat. what is the probability that greg will select a piece of paper with an odd number on it and then select one with an even number on it? \\( \frac{9}{17} \\) \\( \frac{19}{18} \\) \\( \frac{20}{81} \\) \\( \frac{5}{18} \\)

Explanation:

Step1: Identify odd and even numbers

Numbers 1 - 9: odd numbers are 1, 3, 5, 7, 9 (5 numbers); even numbers are 2, 4, 6, 8 (4 numbers).

Step2: Probability of first (odd)

Probability of selecting odd first: $\frac{5}{9}$ (5 odd out of 9 total).

Step3: Probability of second (even, no replacement)

After removing one odd, total left is 8, even numbers still 4. Probability of even second: $\frac{4}{8}=\frac{1}{2}$.

Step4: Multiply probabilities

Since these are dependent events, multiply the probabilities: $\frac{5}{9} \times \frac{4}{8} = \frac{5}{9} \times \frac{1}{2} = \frac{5}{18}$.

Answer:

$\frac{5}{18}$ (the last option, D. $\frac{5}{18}$ assuming the options are labeled A to D with the last being $\frac{5}{18}$)